Consider a neutron star of radius 10 km that spins with a period of 0.8 seconds. imagine a person is standing at the equator of this neutron star. calculate the centripetal acceleration of this person

a = 616850.28 m/s²

Explanation:

Given that,

The radius of the neutron star, r = 10 Km

                                                     = 10,000 m

The time period of the neutron star, T = 0.8 s

The centripetal acceleration is given by the formula,

                                  a = v²/r

The linear velocity is given by the relation,

                                    v = rω

The time taken to complete one complete rotation is given by the relation

                                   T = 2π /ω

Where,

                                    ω = 2π / T

Substituting v and ω into the equation for centripetal acceleration. It becomes

                                    a = 4π²r/T²

Substituting the given values in the above equation

                                      a = 4π² x 10000 / 0.8²

                                         = 616850.28 m/s²

Hence, the centripetal acceleration of this person is, a = 616850.28 m/s²

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.